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Question: How many atoms of sodium (Na) are in \(2.5\)mole of sodium?...

How many atoms of sodium (Na) are in 2.52.5mole of sodium?

Explanation

Solution

We will use the Avogadro number to determine the number of atoms in the given mol. By multiplying the given mole with the Avogadro number we will determine the numbers of atoms. According to the Avogadro, one mole of any substance contains6.02×1023\,6.02 \times {10^{23}} atoms, molecules or ions of that substance.

Complete solution:
According to the Avogadro number, 6.02×1023\,6.02 \times {10^{23}} atoms are present in one mole of any substance and this number,6.02×1023\,6.02 \times {10^{23}} is known as Avogadro number.
According to the Avogadro number,
One mole of any substance =6.02×1023\,6.02 \times {10^{23}} atoms
So,
One mole sodium =6.02×1023\,6.02 \times {10^{23}} atoms of sodium
So, 2.52.5moles of sodium will have,
2.52.5mole of sodium =6.02×1023×2.5\,6.02 \times {10^{23}} \times 2.5 atoms of sodium
=1.5×1024\,1.5 \times {10^{24}} atoms of sodium
So, 1.5×1024\,1.5 \times {10^{24}}atoms of sodium are present in 2.52.5mol of chlorine.

Therefore, 1.5×1024\,1.5 \times {10^{24}} sodium atoms is the correct answer.

Additional information: Most of the time direct moles of the substance are not given. If in place of moles gram amount is given then we have to calculate the moles of that substance first. For this, we use the mole formula. The mole formula is as follows:
mole = massmolarmass{\text{mole}}\,\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,\,{\text{mass}}}}
After calculating the moles we multiply the moles with Avogadro's number to determine the total number of atoms.

Note: The number of atoms present in 12g12\,{\text{g}} of carbon-12 is known as one mole. In the case of monoatomic, one mole of a substance contains an Avogadro number of atoms. Sodium is a monatomic element. We can also determine the total number of molecules and ions by using the Avogadro number. For this, we have to multiply the Avogadro with a total number of molecules or ions.